If α,β are the roots of ax²+bx+c=0 then 1/α³+1/β³=
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we know (a+b)³=a³+3a²b+3ab²+b³
which can be written as (a+b)³=a³+b³+3ab(a+b).
=》a³+b³ = (a+b)³- 3ab(a+b).
similarly
α³+β³= (α+β)³ - 3αβ(α+β).
here The first term is, ax^2 its coefficient is a .
The middle term is, bx its coefficient is b.
The last term, "the constant", is c .
we know in a quadratic equation α+β = -b/a.
and αβ = c/a.
so, substitute these values in (α+β)³ - 3αβ(α+β).
=》(-b/a)³ - 3(c/a)(-b/a).
=》-b³/a³+3bc/a^2.
by taking the LCM OF BOTH..
WE GET.
(-b³+3abc)/a³.
therefore α³+β³ = (-b³+3abc)/a³.
Step-by-step explanation:
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